A capacitor with charge Q is discharged by connecting it to a resistor of resistance R. The graph of the capacitor voltage as a function of time is shown above.

The capacitor is then given the same charge Q but this time is discharged through a resistor of resistance 2R. Which graph represents the capacitor voltage as a function of time now?

1. The original graph (in black)
2. Graph 2 (in red)
3. Graph 3 (in green)
4. Graph 4 (in blue)
5. Graph 5 (in purple)

Because the capacitor starts with the same charge both times the initial voltage is the same. Doubling the resistance doubles the time constant - an increase in the time constant means that any changes take more time, so the capacitor discharges more slowly. The correct graph is the red one, graph 2.

Another way to see this is that the capacitor voltage decreases because charge flows off the capacitor. Increasing the resistance reduces the current, which means the rate at which charge flows off the capacitor is reduced. The capacitor voltage stays high for a longer time.